0=-16(t^2)+72

Solution for 0=-16(t^2)+72 equation:

0=-16(t^2)+72

We move all terms to the left:

0-(-16(t^2)+72)=0

We add all the numbers together, and all the variables

-(-16t^2+72)=0

We get rid of parentheses

16t^2-72=0

a = 16; b = 0; c = -72;
Δ = b2-4ac
Δ = 02-4·16·(-72)
Δ = 4608
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{4608}=\sqrt{2304*2}=\sqrt{2304}*\sqrt{2}=48\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-48\sqrt{2}}{2*16}=\frac{0-48\sqrt{2}}{32}
=-\frac{48\sqrt{2}}{32}
=-\frac{3\sqrt{2}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+48\sqrt{2}}{2*16}=\frac{0+48\sqrt{2}}{32}
=\frac{48\sqrt{2}}{32}
=\frac{3\sqrt{2}}{2}
$